= Solution
On periodic ring states the angular-momentum index $m$ is an integer. At the stated <resonance>,
$$
e^{-i\alpha m^2}=e^{-2\pi im^2}=1,
$$
so \b[$U_T$ is the identity on every orbital state]. After $N$ steps from the normalized ground state,
$$
\psi_N(\theta)=\frac1{\sqrt{2\pi}}e^{iN\eta\cos\theta}.
$$
Its <phase> varies in space but its position <probability density> remains uniform. Periodicity permits <integration by parts>, giving
$$
\langle K\rangle_N=\frac{\hbar^2}{2I}\int_0^{2\pi}
|\partial_\theta\psi_N|^2\,\mathrm d\theta
=\frac{\hbar^2N^2\eta^2}{2I}\frac1{2\pi}
\int_0^{2\pi}\sin^2\theta\,\mathrm d\theta.
$$
Consequently
$$
\boxed{\langle K\rangle_N=\frac{\hbar^2\eta^2}{4I}N^2.}
$$
Equivalently, the angular-momentum <probabilities> are $P_m=J_m(N\eta)^2$, with $\sum_m m^2J_m(z)^2=z^2/2$. The mean <angular momentum> remains zero while its <variance> grows as $N^2$. This <quantum resonance of the kicked rotor> gives coherent ballistic energy growth. The <Bessel second-moment identity> connects the momentum spreading to that quadratic energy law; an incoherent <random walk> would instead give linear energy growth.
\Image[/past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2016/paper-1-kicked-rotor.png]
{title=Original resonant kicked-rotor calculation: angular-momentum probabilities and the quadratic kinetic-energy law}
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